Problem: $\overline{AC}$ is $10$ units long $\overline{BC}$ is $6$ units long $\overline{AB}$ is $2\sqrt{34}$ units long What is $\cos(\angle ABC)$ ? $A$ $C$ $B$ $10$ $6$ $2\sqrt{34}$
Explanation: SOH CAH TOA os = djacent over ypotenuse adjacent $= \overline{BC} = 6$ hypotenuse $= \overline{AB} = 2\sqrt{34}$ $\cos(\angle ABC )=\frac{6}{2\sqrt{34}}$ $=\dfrac{3\sqrt{34} }{34}$